What will be the pH of a solution formed by m | vedclass

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(D) Number of moles of $HCl = \frac{0.1 	imes 40}{1000} = 0.004 \, mol$.Number of moles of $NaOH = \frac{0.45 	imes 10}{1000} = 0.0045 \, mol$.Since

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$12$

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(D) Number of moles of $HCl = \frac{0.1 	imes 40}{1000} = 0.004 \, mol$.Number of moles of $NaOH = \frac{0.45 	imes 10}{1000} = 0.0045 \, mol$.Since $NaOH$ is in excess,the remaining moles of $OH^-$ after neutralization are $0.0045 - 0.004 = 0.0005 \, mol$.The total volume of the solution is $40 + 10 = 50 \, cm^3 = 0.05 \, L$.The concentration of $OH^-$ ions is $[OH^-] = \frac{0.0005 \, mol}{0.05 \, L} = 0.01 \, M = 10^{-2} \, M$.Therefore,$pOH = -\log[OH^-] = -\log(10^{-2}) = 2$.Since $pH + pOH = 14$,we have $pH = 14 - 2 = 12$.

What will be the $pH$ of a solution formed by mixing $40 \, cm^3$ of $0.1 \, M \, HCl$ with $10 \, cm^3$ of $0.45 \, M \, NaOH$?

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